Question

A Carnot's engine has an efficiency of $50\%$ at sink temperature $50{}^{\circ }C$ . Calculate the temperature of source.

• A. $133{}^{\circ }C$
• B. $143{}^{\circ }C$
• C. $100{}^{\circ }C$
• D. $373{}^{\circ }C$

Answer 1

Answer: (D)

$\eta =\frac{\text{ Net work done per cycle }}{\text{ Total amount of heat absorbed per cycle }}$
or $\eta =\frac{W}{Q_1}$
$\therefore$ $\eta =\frac{Q_1-Q_2}{Q_1}$
Or $\eta =1-\frac{Q_2}{Q_1}$
as $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$
$\therefore$ $\eta =1-\frac{T_2}{T_1}$ ..(i)
where, $T_2$ is temperature of sink and $T_1$ is the temperature of the sources.
Here, $\eta =50$%
$T_2=50{}^{\circ }C=273+50$
$=323K$
Putting these values in Eq. (i), we get
$\frac{50}{100}=1-\frac{323}{T_1}$
$T=646K=373{}^{\circ }C$